Question

A condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. Find the energy dissipated in this process.

• A. $\frac{C_2}{C_1+C_2}{U}_0$
• B. $\frac{C_1}{C_1+C_2}{U}_0$
• C. ${\left(\frac{C_1-C_2}{C_1+C_2}\right)}^2{U}_0$
• D. $\frac{C_1{C}_2}{2(C_1+C_2)}{U}_0$

Answer 1

The correct option is A $\frac{C_2}{C_1+C_2}{U}_0$

Here, $U_0=\frac{1}{2}{C}_1{V}_0^2$

After connecting another capacitor, the common potential , $V_c=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1{V}_0+0}{C_1+C_2}=\frac{C_1{V}_0}{C_1+C_2}$

Now energy stored , $U=\frac{1}{2}(C_1+C_2)V_c^2=\frac{1}{2}(C_1+C_2)\frac{C_1^2{V}_0^2}{(C_1+C_2{)}^2}=\frac{C_1{U}_0}{C_1+C_2}$

Energy dissipated $=U_0-U=U_0-\frac{C_1{U}_0}{C_1+C_2}=\frac{C_2{U}_0}{C_1+C_2}$